《复变函数与积分变换(刘建亚)》作业答案.

《复变函数与积分变换》作业参考答案习题1:4、计算下列各式(13i(3i(1+3i-;(323(3i-(513i2z+=,求2z,3z,4z;(761-。解:(13i(3i(1+3i=3i(3+3ii+3=3i(2i+23=6+63i---;(32333(223i3(223i333i41288(3i223i(223i(223i++====++---+;(5213i3i3223i13i4422z++--+===-+,3213i13i131224zzz-++--=⋅=⋅==-,4313i22zzz=⋅=--.(7因为1cosisinππ-=+,所以6221cosisin66kkππππ++-=+,即0k=时,031cosisini6622wππ=+=+;1k=时,133cosisini66wππ=+=;2k=时,25531cosisini6622wππ=+=-+;3k=时,37731cosisini6622wππ=+=--;4k=时,499cosisini66wππ=+=-;5k=时,5111131cosisini6622wππ=+=-.习题2:3、下列函数在何处可导?何处解析?在可导点求出其导数.(22(ifzxy=-;(4(sinchicosshfzxyxy=+(6(azbfzczd+=+。解:(2因为2(,uxyx=,(,vxyy=-,2xux'=,0yu'=,0xv'=,1yv'=-.这四个一阶偏导数都连续,故(,uxy和(,vxy处处可微,但柯西-黎曼方程仅在12x=-上成立,所以(fz只在直线12x=-上可导,此时1122(21xxfzx=-=-'==-,但复平面上处处不解析.(4因为(,sinchuxyxy=,(,cosshvxyxy=,coschxuxy'=,sinshyuxy'=,sinshxvxy'=-,coschyvxy'=.这四个一阶偏导数都连续,故(,uxy和(,vxy处处可微,且满足柯西-黎曼方程,所以(fz在复平面内解析,并且((iiiiiziz(icoschisinshcosisin22cosisincosisin2222cos22yyyyxxyyyyxxyxyxeeeefzuvxyxyxxeeeexxxxeeeeeez-------+-+-'''=+=-=⋅-⋅=-++=⋅+⋅++===.(6020((1(limlim(lim(((zzzfzzfzazzbazbzzczzdczdadbcadbcczczdczdczd∆→∆→∆→⎡⎤+∆-+∆++=-⎢⎥∆∆+∆++⎣⎦--==+∆+++所以,(fz在除dzc=-外处处解析,且2((adbcfzczd-'=+.4、指出下列函数的奇点.(1221(4zzz-+;(2222(1(1zzz+++.解:(122343242242232322(4(1(483448((4(43448(4zzzzzzzzzfzzzzzzzzzz+--+-+-+'==++-+-+=+所以,(fz的奇点为0,2i±.(222232422322(1(12(2(1(213953((1(1(1(1zzzzzzzzzfzzzzz++-+++++++'==-++++所以,(fz的奇点为1-,i±.10、如果(ifzuv=+在区域D内解析,并且满足下列条件之一,试证(fz在D内是一常数.(2(fz在D内解析;证明:由(ifzuv=+在区域D内解析,知(,uxy、(,vxy在区域D内可微,且xyuv''=,yxuv''=-.同理,由(fz在D内解析,知xyuv''=-,yxuv''=.从而我们得到0xyyxuvuv''''====,所以(,uxy、(,vxy皆为常数,故(fz在D内是一常数.15、求解下列方程:(210ze+=解:1ze=-,于是Ln(1ln1iarg(12i=(21i,zkkkZππ=-=+-++∈18、求Ln(i-,Ln(34i-+的值及主值.解:Ln(ilniiarg(i2ii2i2kkπππ-=-+-+=-+,所以其主值为i2π-;4Ln(34iln34iiarg(34i2iln5i(arctan2i3kkπππ-+=-++-++=+-+,所以其主值为4ln5i(arctan3π+-.19、求1i2eπ-,1i4eπ+,i3,i(1i+的值.解:1ii(22cos(isin(i22eeeeeππππ--⎡⎤=⋅=-+-=-⎢⎥⎣⎦;(1i11i444444222cosisini1i44222eeeeeeππππ+⎛⎫⎛⎫=⋅=+=+=+⎪⎪⎪⎝⎭⎝⎭;(iiLn3i(ln32i2+iln323cosln3isinln3kkkeeeeπππ+--====+;11iln2i2i2iln22iiln(1i444ln2ln2(1icosisin22kkkeeeeππππ⎛⎫⎛⎫⎛⎫++-++-+⎪⎪⎪+⎝⎭⎝⎭⎝⎭⎛⎫+====+⎪⎝⎭.20、求21,2(2-,i1-,ii,1i(34i+-的值.解:22Ln122i1cos(22isin(22keekkπππ===+;(22Ln(22ln2(212i2(22cos(212isin(212keekkπππ-++⎡⎤⎡⎤-===+++⎣⎦⎣⎦;iiLn1i(2i21kkeeeππ---===;1ii2i2iiLni22ikkeeeπππ⎛⎫⎛⎫+-+⎪⎪⎝⎭⎝⎭===;((444(1iln5arctani2iln5arctan2iln5arctan231i(1iLn(34i332(34i45cosln5isinln5,arctan,3kkkkeeeekZπππθπθθθ⎡⎤⎛⎫⎛⎫++-++-+-+⎪⎢⎥⎪++-⎝⎭⎣⎦⎝⎭--====-+-=∈⎡⎤⎣⎦22、解方程:(1ch0z=;解:1Arch0Ln(001Lni2i2zkπ⎛⎫==+-==+⎪⎝⎭,kZ∈.习题3:1、沿下列路径计算积分2i20zdz+⎰:(1从原点至2i+的直线段;(2从原点沿实轴至2,再由2铅直向上至2i+;(3从原点沿虚轴至i,再由i沿水平方向向右至2i+.解:(1从原点至2i+的直线段的复参数方程为i2xzx=+,1(1i2dzdx=+,参数:02x→,所以22i22323330001111(1i(1i(2i2323zdzxdxx+=+=+=+⎰⎰(2从原点沿实轴至2的直线段的复参数方程为zx=,参数:02x→,由2铅直向上至2i+的直线段的复参数方程为2izy=+,参数:01y→,所以122i212222202132300(2ii18i2111(i44i24i=i(2i333333CCzdzzdzzdzxdxydyxyydy+=+=++=+--+=--++=+⎰⎰⎰⎰⎰⎰(3从原点沿虚轴至i的直线段的复参数方程为izy=,参数:01y→,由i沿水平方向向右至2i+的复参数方程为izx=+,参数:02x→,所以122i1222222012223300(ii(ii1i1i(i(2i(2i3333CCzdzzdzzdzydyxdxydyxdx+=+=++=-++=-+++=+⎰⎰⎰⎰⎰⎰⎰2、分别沿yx=与2yx=算出积分1i20(ixydz+-⎰的值.解:yx=的复参数方程为(1izx=+,(1idzdx=+,参数:01x→所以1i122051(i(i(1ii66xydzxxdx+-=-+=-⎰⎰;2yx=的复参数方程为2izxx=+,(12idzxdx=+,参数:01x→所以1i1222051(i(i(12ii66xydzxxxdx+-=-+=+⎰⎰5、计算积分Czdzz⎰的值,其中C为正向圆周:(13z=解:设1C是C内以被积函数的奇点0z=为圆心的正向圆周,那么111132i=6iCCCCzzzzdzdzdzzdzzzzzzππ⋅====⋅⎰⎰⎰⎰6、试用观察法得出下列积分的值,并说明观察时所依据的是什么?C是正向圆周1z=:(12Cdzz+⎰;(2223Cdzzz++⎰;(3cosCdzz⎰;(413Cdzz-⎰;(5zCzedz⎰;(6i522Cdzzz⎛⎫⎛⎫++⎪⎪⎝⎭⎝⎭⎰.解:(102Cdzz=+⎰,根据柯西积分定理;(22023Cdzzz=++⎰,根据柯西积分定理;(30cosCdzz=⎰,根据柯西积分定理;(42i13Cdzzπ=-⎰,根据复合闭路定理;(50zCzedz=⎰,根据柯西积分定理;(64ii55i22Cdzzzπ=-⎛⎫⎛⎫++⎪⎪⎝⎭⎝⎭⎰,根据柯西积分定理及复合闭路定理.7、沿指定曲线的正向计算下列积分:(13zCedzz-⎰,:31Cz-=;(222Cdzza-⎰,:Czaa+=;(3i21zCedzz+⎰,4:2i3Cz-=;(43Czdzz+⎰,:2Cz=;(523(1(1Cdzzz+-⎰,:1Czr=;(63cosCzzdz⎰,C为包围0z=的闭曲线;(722(1(4Cdzzz+-⎰,3:2Cz=;(8sinCzdzz⎰,:3Cz=;(92cos2Czdzzπ⎛⎫-⎪⎝⎭⎰,:3Cz=;(105zCedzz⎰,:1Cz=.解:(1332i2i3zzzCedzeezππ==⋅=-⎰;(22212iiCzadzzazaaππ=-=⋅=---⎰;(3ii2i2i1izzCzeedzzzeππ==⋅=++⎰;(403Czdzz=+⎰;(5230(1(1Cdzzz=+-⎰;(63cos0Czzdz=⎰;(7222222ii1(1(42i(i(4(i(411102i44CCCzzdzdzdzzzzzzzzz=-=⎡⎤-=-⎢⎥+-+---⎣⎦⎛⎫-=-=⎪--⎝⎭⎰⎰⎰;(80sin2isin0zCzdzzzπ==⋅=⎰;(9(22cos2sin21!2Czzidzzizππππ==⋅-=-⎛⎫-⎪⎝⎭⎰;(10502(51!12zzCzeiidzezππ==⋅=-⎰.21、证明:22uxy=-和22yvxy=+都是调和函数,但是iuv+不是解析函数.证明:因为2uxx∂=∂,222ux∂=∂,2uyy∂=-∂,222uy∂=-∂,2222(vxyxxy∂-=∂+,223222362(vxyyxxy∂-=∂+,22222(vxyyxy∂-=∂+,232222326(vyxyyxy∂-=∂+,所以22220uuxy∂∂+=∂∂,22220vvxy∂∂+=∂∂,且xyuv''≠,yxuv''≠-.即22uxy=-和22yvxy=+都是调和函数,但是iuv+不是解析函数.22、由下列各已知调和函数求解析函数(ifzuv=+,并写出z的表达式:(122((4uxyxxyy=-++;(222yvxy=+,(20f=;(32(1uxy=-,(2if=-.解:(1因为(ifzuv=+是调和函数,所以22363vuxxyyxy∂∂=-=-++∂∂,22363vuxxyyyx∂∂==+-∂∂.于是22223(363(33vxxyydygxxyxyy=+-=++-⎰.那么222(63363vgxxyyxxyyx∂'=++=-++∂,则3(gxxC=-+,所以322333vxxyxyyC=-++-+,3223322332233((33i(33i(1i3(i3(i(ii(1iifzxxyxyyxxyxyyCxxyxyyCzC=+--+-++-+⎡⎤=-++++⎣⎦=-+(22222(vxyxxy∂-=∂+,22222(vxyyxy∂-=∂+.因为(ifzuv=+是调和函数,所以222222222222222(i11(ii((((iyxxyxyxyfzvvxyxyxyxyz---'''=+=+===++++,从而1(fzCz=-+.由(20f=知12C=,所以11(2fzz=-.(3因为(ifzuv=+是调和函数,所以2(1vuxxy∂∂=-=--∂∂,2vuyyx∂∂==∂∂.于是22(vydygxy==+⎰.那么(2(1vgxxx∂'==--∂,则2(2gxxxC=-++,所以222vxxyC=-+++,2222((22i(2ii(i2(i1ii(1ifzxyyxxyCxyxyCzC=-+-+++⎡⎤=-+-+++⎣⎦=-